Algebra Worksheets Polynomial Functions Worksheets Rational - Free Printable

To solve the problem of listing all possible zeroes for each polynomial using the Rational Zero Theorem, we need to follow these steps:

Rational Zero Theorem

The Rational Zero Theorem states that if a polynomial
$$
P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0
$$
has integer coefficients, then every rational zero of the polynomial is of the form:
$$
\frac{p}{q}
$$
where:
- \( p \) is a factor of the constant term \( a_0 \),
- \( q \) is a factor of the leading coefficient \( a_n \).

Steps to Solve

1. Identify the constant term \( a_0 \) and list all its factors.
2. Identify the leading coefficient \( a_n \) and list all its factors.
3. Form all possible rational zeros by taking the ratio \( \frac{p}{q} \), where \( p \) is a factor of \( a_0 \) and \( q \) is a factor of \( a_n \).

Solving Each Polynomial

#### Problem 1: \( x^3 + 14x^2 - 32x - 15 \)
- Constant term \( a_0 = -15 \): Factors are \( \pm 1, \pm 3, \pm 5, \pm 15 \).
- Leading coefficient \( a_n = 1 \): Factors are \( \pm 1 \).
- Possible rational zeros:
$$
\pm 1, \pm 3, \pm 5, \pm 15
$$

#### Problem 2: \( x^3 - 13x^2 - 27x + 24 \)
- Constant term \( a_0 = 24 \): Factors are \( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24 \).
- Leading coefficient \( a_n = 1 \): Factors are \( \pm 1 \).
- Possible rational zeros:
$$
\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24
$$

#### Problem 3: \( x^4 - 3x^3 + 5x^2 - 12x + 36 \)
- Constant term \( a_0 = 36 \): Factors are \( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 9, \pm 12, \pm 18, \pm 36 \).
- Leading coefficient \( a_n = 1 \): Factors are \( \pm 1 \).
- Possible rational zeros:
$$
\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 9, \pm 12, \pm 18, \pm 36
$$

#### Problem 4: \( x^4 + 11x^3 - 23x^2 + x - 18 \)
- Constant term \( a_0 = -18 \): Factors are \( \pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18 \).
- Leading coefficient \( a_n = 1 \): Factors are \( \pm 1 \).
- Possible rational zeros:
$$
\pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18
$$

#### Problem 5: \( 2x^3 + 17x^2 - 36x - 8 \)
- Constant term \( a_0 = -8 \): Factors are \( \pm 1, \pm 2, \pm 4, \pm 8 \).
- Leading coefficient \( a_n = 2 \): Factors are \( \pm 1, \pm 2 \).
- Possible rational zeros:
$$
\pm 1, \pm 2, \pm 4, \pm 8, \pm \frac{1}{2}, \pm \frac{2}{2} = \pm 1, \pm \frac{4}{2} = \pm 2, \pm \frac{8}{2} = \pm 4
$$
Simplifying, the unique values are:
$$
\pm 1, \pm 2, \pm 4, \pm 8, \pm \frac{1}{2}
$$

#### Problem 6: \( 3x^3 - 24x^2 + 16x - 12 \)
- Constant term \( a_0 = -12 \): Factors are \( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12 \).
- Leading coefficient \( a_n = 3 \): Factors are \( \pm 1, \pm 3 \).
- Possible rational zeros:
$$
\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{4}{3}, \pm \frac{6}{3} = \pm 2, \pm \frac{12}{3} = \pm 4
$$
Simplifying, the unique values are:
$$
\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{4}{3}
$$

#### Problem 7: \( 4x^3 - 31x^2 - 14x + 16 \)
- Constant term \( a_0 = 16 \): Factors are \( \pm 1, \pm 2, \pm 4, \pm 8, \pm 16 \).
- Leading coefficient \( a_n = 4 \): Factors are \( \pm 1, \pm 2, \pm 4 \).
- Possible rational zeros:
$$
\pm 1, \pm 2, \pm 4, \pm 8, \pm 16, \pm \frac{1}{2}, \pm \frac{2}{2} = \pm 1, \pm \frac{4}{2} = \pm 2, \pm \frac{8}{2} = \pm 4, \pm \frac{16}{2} = \pm 8, \pm \frac{1}{4}, \pm \frac{2}{4} = \pm \frac{1}{2}, \pm \frac{4}{4} = \pm 1, \pm \frac{8}{4} = \pm 2, \pm \frac{16}{4} = \pm 4
$$
Simplifying, the unique values are:
$$
\pm 1, \pm 2, \pm 4, \pm 8, \pm 16, \pm \frac{1}{2}, \pm \frac{1}{4}
$$

#### Problem 8: \( 5x^3 + 16x^2 - 7x + 30 \)
- Constant term \( a_0 = 30 \): Factors are \( \pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 30 \).
- Leading coefficient \( a_n = 5 \): Factors are \( \pm 1, \pm 5 \).
- Possible rational zeros:
$$
\pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 30, \pm \frac{1}{5}, \pm \frac{2}{5}, \pm \frac{3}{5}, \pm \frac{6}{5}, \pm \frac{10}{5} = \pm 2, \pm \frac{15}{5} = \pm 3, \pm \frac{30}{5} = \pm 6
$$
Simplifying, the unique values are:
$$
\pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 30, \pm \frac{1}{5}, \pm \frac{2}{5}, \pm \frac{3}{5}, \pm \frac{6}{5}
$$

Final Answer

Here are the possible rational zeros for the selected problems:

1. \( x^3 + 14x^2 - 32x - 15 \):
$$
\boxed{\pm 1, \pm 3, \pm 5, \pm 15}
$$

2. \( x^3 - 13x^2 - 27x + 24 \):
$$
\boxed{\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24}
$$

3. \( x^4 - 3x^3 + 5x^2 - 12x + 36 \):
$$
\boxed{\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 9, \pm 12, \pm 18, \pm 36}
$$

4. \( x^4 + 11x^3 - 23x^2 + x - 18 \):
$$
\boxed{\pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18}
$$

5. \( 2x^3 + 17x^2 - 36x - 8 \):
$$
\boxed{\pm 1, \pm 2, \pm 4, \pm 8, \pm \frac{1}{2}}
$$

6. \( 3x^3 - 24x^2 + 16x - 12 \):
$$
\boxed{\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{4}{3}}
$$

7. \( 4x^3 - 31x^2 - 14x + 16 \):
$$
\boxed{\pm 1, \pm 2, \pm 4, \pm 8, \pm 16, \pm \frac{1}{2}, \pm \frac{1}{4}}
$$

8. \( 5x^3 + 16x^2 - 7x + 30 \):
$$
\boxed{\pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 30, \pm \frac{1}{5}, \pm \frac{2}{5}, \pm \frac{3}{5}, \pm \frac{6}{5}}
$$